PERMUTATIONSPERMUTATIONSandand

COMBINATIONSCOMBINATIONSPERLIS PERLIS

MATRICULATION MATRICULATION COLLEGECOLLEGE

QS 026QS 026CHAPTER 9CHAPTER 9

Permutations of a set of objectsPermutations of a set of objects

A permutation of a set of objects is any arrangement of those objects in a definite order.

(order is important).

For example, if A={a,b,c,d} , then abab two-element permutation of A, acdacd three-element permutation of A, adcbadcb four-element permutation of A.

The order in which objects are arranged is The order in which objects are arranged is important. important. For example,For example, abab and and baba are considered different two-are considered different two- element permutationselement permutations abc abc andand cba cba are distinct three-element are distinct three-element permutations, and permutations, and

abcdabcd and and cbad cbad are different four-element are different four-element permutations.permutations.

For another example the six permutations of For another example the six permutations of ABC are the six different arrangements of ABC. ABC are the six different arrangements of ABC. These areThese are

ABC ACB BAC BCA CAB CBA

If there are 4 ways from If there are 4 ways from Johor to Penang and 2 Johor to Penang and 2 ways from Penang to ways from Penang to Pulau Langkawi, how Pulau Langkawi, how many ways can we go for many ways can we go for a journey from Johor to a journey from Johor to Pulau Langkawi through Pulau Langkawi through Penang.Penang.

Bus   Taxi Flight  Johor Train Penang Ferry P.Langkawi   Van

The number of permutations can be calculated The number of permutations can be calculated using the using the multiplication principlemultiplication principle..

Multiplication PrincipleMultiplication Principle

If there are If there are m m ways for an event to occur and ways for an event to occur and nn ways for another event to occur, then there are ways for another event to occur, then there are mm x x nn ways for the two events to occur. ways for the two events to occur.

ExampleExampleA coin and a dice are tossed together. How A coin and a dice are tossed together. How many different outcomes are possible ? many different outcomes are possible ?

2 6 12

The possible outcomes are

(H,1) (H,2) (H,3)

(T,2) (T,3) (T,5)

(H,4)

(T,4)(T,1)

(H,6)(H,5)

(T,6)

SolutionThe coin has two possible outcomes (head, H and Tail, T) and the dice has 6 possible outcomes.

The number of different possible outcomes is ___ x ___ = ____

Die

1 2 3 4 5 6

Coin

Head (H)

(H,1) (H ,2) (H, 3) (H, 4) (H, 5) (H, 6)

Tail (T) (T,1) (T,2) (T, 3) (T, 4) (T, 5) (T, 6)

ExampleExample

A shop stocks T-shirts in four sizes : small, medium A shop stocks T-shirts in four sizes : small, medium and large. They are available in four colours; black , and large. They are available in four colours; black , red , yellow and green. If the sizes are denoted by S, red , yellow and green. If the sizes are denoted by S, M and L and the colours are denoted by B, R, Y and M and L and the colours are denoted by B, R, Y and G make a list of all the different labels needed to G make a list of all the different labels needed to distinguish the T-shirts and find the number of distinguish the T-shirts and find the number of different labels. different labels.

Solution

SBSB SRSR SYSY SGSG MBMB MRMRMYMY MGMG LBLB LRLR LYLY LGLG

The number of different labels is 3 x 4 = 12

Permutations of n objectsPermutations of n objects

We will now consider the method for finding a We will now consider the method for finding a number of permutations on the letters A, B and C number of permutations on the letters A, B and C using the multiplication principle. How many using the multiplication principle. How many arrangements of the letters A, B and C are there?arrangements of the letters A, B and C are there?

Let us consider the number of ways of arranging n Let us consider the number of ways of arranging n letter.letter.

If we have 1 letter, there is just one arrangement. If we have 1 letter, there is just one arrangement. Example : AExample : A

If we have 2 letters, there are two different arrangements. Example : AB and BA

there are three ways of choosing the first letter.

If we have 3 letters, the different arrangements are :

C

or

B

or

A

When the first letter has been chosen, there are two letters When the first letter has been chosen, there are two letters from which to choose the second; and the possible ways of from which to choose the second; and the possible ways of choosing the first two letters are: choosing the first two letters are:

there are two ways of choosing the second letter

Cor

Bor

A

B

or

A

C

or

A

C

or

B

i.e. for each of the three ways of choosing the i.e. for each of the three ways of choosing the first letter, there are two ways of choosing the first letter, there are two ways of choosing the second letter. second letter.

Hence there are 3 x 2 ways of choosing the first Hence there are 3 x 2 ways of choosing the first two letters.two letters.

Having chosen the first two letters, there is only Having chosen the first two letters, there is only one choice for the third letter, i.e. for each of the one choice for the third letter, i.e. for each of the 3 x 2 ways of choosing the first two letters, there 3 x 2 ways of choosing the first two letters, there is only one possibility for the third letter. is only one possibility for the third letter.

Hence there are 3 x 2 x 1 ways of arranging the Hence there are 3 x 2 x 1 ways of arranging the three letters A, B and C.three letters A, B and C.

NoteNote : if repetition are allowed, we can choose from all 3 : if repetition are allowed, we can choose from all 3 digits for each digit of the number. A digit can be used digits for each digit of the number. A digit can be used more than once.more than once.

How many different ways of arranging 3 digit numbers from digits 5 and 6 ?

5

6

5

6

5

6

5

65

65

65

6there will be 8 different ways, which is found from 2 2 2

How many different ways do you think there How many different ways do you think there are of arranging 4 letters?are of arranging 4 letters?

You should able to see, there will be 24 different ways, which is found from 4 x 3 x 2 x 2 x 1.

If there are 500 different objects, the number of ways would be 500 x 499 x 498 x … x 3 x 2 x 1.

This is tedious to write, so we use the notation 500! ( factorial 500 ).

ExampleExample

List the set of all permutations of the symbols List the set of all permutations of the symbols P, Q and R when they are taken 3 at a timeP, Q and R when they are taken 3 at a time

Solution

PQR, PRQ, QPR, QRP, RPQ, RQP

Number of permutations of n different objects taken all at a time without repetition

nnPP

nn= n x (n – 1) x(n – 2) x … x 2 x 1= n!

In general,

Pn

n = n!

NotesNotes : n! n! means the products of all the means the products of all the integers from 1 to n inclusive and is integers from 1 to n inclusive and is

called ‘n factorial’. called ‘n factorial’.

Example Example

How many three-digit numbers can be made How many three-digit numbers can be made from the integers 2, 3, 4 ?from the integers 2, 3, 4 ?

SolutionSolutionn = 3

P P3 = 3 ! 3 x 2 x 1 = 6

The number of arrangements is 6.

n 3=n

Example Example

In how many ways can ten instructors be In how many ways can ten instructors be assigned to ten sections of a course in assigned to ten sections of a course in mathematics?mathematics?

Pn P10n 10

= 10 ! = 3,628,800 ways

Solution

Substituting n = 10 we get

=

Example Example

Three people, Aishah, Badrul and Daniel must be scheduled Three people, Aishah, Badrul and Daniel must be scheduled for job interviews. In how many different orders can this be for job interviews. In how many different orders can this be done?done?

Solution

n = 3

So there are 3! = 6 possible orders for the interviews.

Example Example

How many different numbers can be formedHow many different numbers can be formedfrom the digits 5, 6, 7 and 8from the digits 5, 6, 7 and 8

i)   if no repetitions are allowedi)   if no repetitions are allowed n = 4 npn = 4p4 = 4! = 24 ii)  if the first digit must be 7. 1p1 X 3p3 = 1! X 3! = 6